Mixed Review

KEY Chemistry: Mixed Review

Solve each of the following problems, being sure to show your work and include all proper units.

1. Use the ideal gas law PV = nRT, to derive Boyle’s law and Charles’s law.

If n, R and T are constant: PV = nRT If n, R and P are constant: V/T = nRP PV = constant V/T = constant

P1V1 = P2V2 (Boyle’s Law) V1/T1 = V2/T2 (Charles’s Law)

2. A container holds 265 mL of chlorine gas, Cl2. Assuming that the gas sample is at STP, what is its mass?

PV (1 atm)(0.265 L) V = 265 mL (0.265 L) PV = nRT n n T = 273 K RT (0.0821 L atm/mol K)(273 K) P = 1 atm n = ? mol . . n 0.0118 mol Cl R = 0.0821 L atm / mol K 2

71 g Cl2 x g Cl2 0.0118 mol Cl2 0.84 g Cl2 1 mol Cl2

3. Suppose that 3.11 mol of carbon dioxide is at a pressure of 0.820 atm and a temperature of 39oC. What is the volume of the sample, in liters? nRT n = 3.11 mol CO2 PV = nRT V P = 0.820 atm P T = 39oC + 273 = 312 K (3.11 mol)(0.082 1 L atm/mol K )(312 K) V = ? L V R = 0.0821 L.atm / mol.K 0.820 atm

V 97.1 L

4. Compare the rates of diffusion of carbon monoxide, CO, and sulfur trioxide, SO3.

v 80 g v 1.69 1 1 (or SO is 0.59x slower) CO diffuses 1.69x faster than SO3 3 v2 28 g v2 1

5. A gas sample that has a mass of 0.993 g occupies 0.570 L. Given the temperature is 281 K and the pressure is 1.44 atm, what is the molar mass of the gas?

PV (1.44 atm)(0.570 L) P = 1.44 atm PV = nRT n n V = 0.570 L RT (0.0821 L atm/mol K)(281 K) n = ? mol . . R = 0.0821 L atm / mol K n 0.0356 mol T = 281 K

g 0.993 g molar mass 28 g/mol mol 0.0356 mol

Answers: 1. 2. 0.84 g Cl2 3. 97.1L 4. CO is 1.69x faster 5. 28 g/mol

6. The density of a gas is 3.07 g/L at STP. Calculate the gas’s molar mass.

P = 1 atm mass PV (1 atm)(0.325 7 L) D n n V = ? L volume RT (0.0821 L atm/mol K)(273 K) T = 273 K M n = ? mol 3.07g/L M 68.8 g n 0.01453 mol 22.4 L R = 0.0821 M M g 1.0 g D = 3.07 g/L D 3.07 g/L molar mass V V mol 0.01453 mol Density is an intensive property. 1.0 g Assume mass = 1.0 g V 0.3257 L 3.07 g/L molar mass 68.8 g/mol

7. How many moles of helium gas would it take to fill a gas balloon with a volume of 1000. cm3 when the temperature is 32oC and the atmospheric pressure is 752 mm Hg? 1 atm P = 752 mm Hg PV = nRT 752 mm Hg 0.9895 atm 760 mm Hg V = 1 L PV n = ? mol n (0.9895 atm)(1 L) RT n . . (0.0821 L atm/mol K)(305 K) R = 0.0821 L atm / mol K n 0.0395 mol T = 32oC + 273 = 305 K

8. A gas sample is collected at 16oC and 0.982 atm. If the sample has a mass of 7.40 g and a volume of 3.96 L, find the volume of the gas at STP and the molar mass.

P V P V (0.982 atm)(3.96 L) (1 atm)(V ) 1 1 2 2 2 P1 = 0.982 atm V 3.67 L T1 T2 289 K 273 K 2 V1 = 3.96 L o T1 = 16 C + 273 = 289 K P2 = 1 atm (0.982 atm)(3.96 L) V = ? L PV = nRT n 2 (0.0821 L atm/mol K)(289 K) n 0.164 mol T2 = 273 K g 7.4 g molar mass molar mass 45 g/mol mol 0.164 mol 9. An unknown gas effuses at 0.850 times the effusion rate of nitrogen dioxide, NO2. Estimate the molar mass of the unknown gas.

Unknown Gas Nitrogen dioxide, NO2 v m 0.850 46 g m = ? g m = 46 g 1 2 1 2 v m 1 x g v1 = 0.850 v2 = 1 2 1

V 63.7 g 2

10. What will be the pressure if we open: a. just valve A? 0.75 atm

b. just valve C? 0.5 atm

c. any two valves? 0.5 atm

Use Boyle’s law: (a) 1.5 atm of gas divided in 2 spheres = 0.75 atm (b) 1.0 atm of gas divided in 2 spheres = 0.5 atm (c) 1.5 atm of gas divided in 3 spheres = 0.5 atm

Answers: 6. 68.8 g/mol 7. 0.0395 mol 8. 3.67 L & 45 g/mol 9. 63.7 g/mol